The continued fraction expansion for the number e = 2.718281828459045.....

is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

How many terms are required to get 4 decimal places (2.718) correct?

Please teach me how to solve this question. Thank you very much.

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- October 16th 2006, 10:01 AMbeta12The continued fraction expansion for e
The continued fraction expansion for the number e = 2.718281828459045.....

is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

How many terms are required to get 4 decimal places (2.718) correct?

Please teach me how to solve this question. Thank you very much. - October 16th 2006, 05:14 PMThePerfectHacker
That is a classical expansion by Euler (sadly not even I know the proof).

But I hope you are asking not to prove it but to get 4 decimals?

Using a well-known inequlality from continued fractions:

|x-p_n/q_n|<=(1/q_n)^2

In order to get 4 decimal points we require that,

(1/q_n)^2<=.0001

Equivalently,

1/q_n<=.0316

Thus,

q_n>=31.62

Thus,

q_n>=32

That is the smallest "n" that makes the denominator of the convergent at least 32. - October 17th 2006, 05:22 AMbeta12
Hi Perfecthacker,

Thank you very much.